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9.6 Other Heat Conduction
Problems
We discussed in 9.5, the heat conduction
equation α2uxx = ut , 0 < x < L, t > 0, (1)
the boundary conditions u(0,t)=0, u(L,t) =0, t >
0, (2) and the initial condition u(x,0)=f(x), 0 ≤
x ≤ L. (3)
 We now consider two other problems of onedimensional heat conduction that can be
handled by the method developed in Section
9.5.
1. Nonhomogeneous Boundary Conditions
2. Bar with Insulated Ends

Nonhomogeneous Boundary
Conditions

Suppose now that one end of the bar
is held at a constant temperature T1
and the other is maintained at a
constant temperature T2. Then the
boundary conditions are
u(0,t) = T1, u(L,t) = T2, t > 0. (4)
The differential equation (1) and the
initial condition (3) remain unchanged.
The Steady State Solution.
After a long time—that is, as t →∞—we anticipate
that a steady temperature distribution v(x) will be
reached, which is independent of the time t and the
initial conditions. Since v(x) must satisfy the
equation of heat conduction (1), we have
v''(x) = 0, 0 < x < L. (5)
Hence the steady-state temperature distribution is a
linear function of x. Further v(x) must satisfy the
boundary conditions
v(0) = T1, v(L) = T2, (6)
which are valid even as t →∞. The solution of Eq. (5)
satisfying Eqs. (6) is
v(x) = (T2 − T1)x/L + T1. (7)

The Transient Solution.
Returning to the original problem, Eqs. (1),
(3), and (4), we will try to express u(x, t) as
the sum of the steady-state temperature
distribution v(x) and another (transient)
temperature distribution w(x, t); thus we write
u(x, t) = v(x) + w(x, t). (8)
We obtain

where
Example

Consider the heat conduction problem
uxx = ut , 0 < x < 30, t > 0, u(0, t) = 20, u(30, t) = 50, t >
0, u(x, 0) = 60 − 2x, 0 < x < 30. Find the steady-state
temperature distribution and the boundary value
problem that determines the transient distribution.
Bar with Insulated Ends
A slightly different problem occurs if
the ends of the bar are insulated so
that there is no passage of heat
through them. That is,
α2uxx = ut , 0 < x < L, t > 0,
the boundary conditions
ux (0, t) = 0, ux (L, t) = 0, t > 0,
and the initial condition
u(x,0)=f(x), 0 ≤ x ≤ L.

Bar with Insulated Ends

1. Solve by the method of separation of
variables to get the fundamental solutions for
the problem

2. Solution to the heat conduction problem for
a rod with insulated ends,
Example

Find the temperature u(x, t) in a metal rod of length 25
cm that is insulated on the ends as well as on the sides
and whose initial temperature distribution is u(x, 0) = x
for 0 < x < 25.
More General Problems
The method of separation of variables
can also be used to solve heat
conduction problems with other
boundary conditions than those given
by previous two types.
 For example, the left end of the bar
might be held at a fixed temperature
T, while the other end is insulated. In
this case the boundary conditions are
u(0, t) = T, ux (L, t) = 0, t > 0.

9.7 The Wave Equation: Vibrations
of an Elastic String
a2uxx = utt
Wave Equation
Let u(x, t) denote the vertical displacement experienced
by the string at the point x at time t. If damping
effects, such as air resistance, are neglected, and if
the amplitude of the motion is not too large, then u(x,
t) satisfies the partial differential equation a2uxx = utt
in the domain 0 < x < L, t > 0 where a is the velocity
of propagation of waves along the string given by a2
= T/ρ for T is the tension (force) in the string, and ρ is
the mass per unit length of the string material.
Boundary and Initial Conditions
The ends are assumed to remain fixed, and
therefore the boundary conditions are
u(0,t) = 0, u(L,t) = 0, t ≥ 0.
 Initial Conditions:
The initial position of the string
u(x, 0) = f (x), 0 ≤ x ≤ L
and its initial velocity
ut (x, 0) = g(x), 0 ≤ x ≤ L,
where f and g are given functions. For
consistency, f (0) = f (L) = 0, g(0) = g(L) = 0.

Example: Elastic String with
Nonzero Initial Displacement.
First suppose that the string is disturbed from its
equilibrium position and then released at time t = 0
with zero velocity to vibrate freely.
Then the vertical displacement u(x, t) must satisfy the
wave equation
a2uxx = utt, 0 < x < L, t > 0; (1)
the boundary conditions
u(0, t) = 0, u(L, t) = 0, t ≥ 0; (3)
and the initial conditions
u(x, 0) = f (x), ut (x, 0) = 0, 0 ≤ x ≤ L, (9)
where f is a given function describing the
configuration of the string at t = 0.

Solve by method of separation of
variables
The formal solution of the problem is
Where

The quantities nπa/L for n = 1, 2, . . . are the natural
frequencies of the string. The factor sin (nπx/L)
represents the displacement pattern occurring in the
string when it is executing vibrations of the given
frequency. Each displacement pattern is called a
natural mode of vibration and is periodic in the space
variable x; the spatial period 2L/n is called the
wavelength of the mode of frequency nπa/L.
Example

Consider a vibrating string of length L = 30
that satisfies the wave equation 4uxx = utt,
0 < x < 30, t > 0. Assume that the ends of
the string are fixed and that the string is set
in motion with no initial velocity from the
initial position
u(x, 0) = f (x) ={x/10, 0 ≤ x ≤ 10,
{(30 − x)/20, 10 < x ≤ 30.
Find the displacement u(x, t) of the string
and describe its motion through one period.
Answer
where
Plots of u versus x for fixed values
of t for the string
Plots of u versus t for fixed values
of x for the string
General Problem for the Elastic
String.
Let us modify the problem just considered
by supposing that the string is set in motion
from its equilibrium position with a given
velocity. Then the vertical displacement
u(x,t) must satisfy the wave equation (1)
a2uxx = utt, 0 < x < L, t > 0;
the boundary conditions (3)
u(0, t) = 0, u(L, t) = 0, t ≥ 0;
and the initial conditions
u(x, 0) = 0, ut (x, 0) = g(x), 0 ≤ x ≤ L, (31)
where g(x) is the initial velocity at the point x
of the string.

Formal solution
Where
Use of the principle of superposition.
 To solve the wave equation with the general
initial conditions u(x,0) = f(x), ut(x,0)=g(x),
0<x<L, you can solve instead the somewhat
simpler problems with the initial conditions (9)
and (31) discussed before, respectively, and
then add together the two solutions.
9.8 Laplace’s Equation




Laplace’s equation in two dimensions is uxx + uyy =
0, and in three dimensions uxx + uyy + uzz = 0.
The potential function of a particle in free space
acted on only by gravitational forces satisfies the
same equations. So, Laplace’s equation is also
called potential equation.
The problem of finding a solution of Laplace’s
equation that takes on given boundary values is
known as a Dirichlet problem (the first boundary
value problem of potential theory).
If the values of the normal derivative are prescribed
on the boundary, the problem is said to be a
Neumann problem (the second boundary value
problem of potential theory).
Dirichlet Problem for a Rectangle
The mathematical problem of finding the
function u satisfying Laplace’s equation (1)
uxx + uyy = 0, in the rectangle 0 < x < a, 0 < y <
b, and also satisfying the boundary conditions
u(x, 0) = 0, u(x, b) = 0, 0 < x < a,
u(0, y) = 0, u(a, y) = f (y), 0 ≤ y ≤ b
where f is a given function on 0 ≤ y ≤ b.
Solution
To solve this problem, we construct a
fundamental set of solutions satisfying the
partial differential equation and the
homogeneous boundary conditions; then we
superpose these solutions so as to satisfy the
remaining boundary condition.
 Solution

where
Example
In above problem,
let a = 3, b = 2, and
f(y)={ y, 0 ≤ y ≤ 1,
{2 − y, 1 ≤ y ≤ 2.
Answer
We find that
Plots and level curves
are in the figures.
Dirichlet Problem for a Circle

Consider the problem of solving Laplace’s
equation in a circular region r < a subject to
the boundary condition u(a, θ) = f (θ), where
f is a given function on 0 ≤ θ < 2π (see
Figure). In polar coordinates Laplace’s
equation has the form
u must be periodic in θ
with period 2π. Moreover
u(r, θ) is bounded for r ≤ a.
Apply the method of separation of
variables to solve this problem

Above equation represents the solution of the
boundary value problem
where
Chapter Summary
Section 9.2 Fourier Series & Section 9.3
The Fourier Convergence Theorem
Section 9.4 Even and Odd
Functions
Section 9.5 Separation of Variables;
Heat Conduction in a Rod Section
9.6 Other Heat Conduction Problems
The classical equation for unidirectional
heat conduction in a rod of finite length L is
α2uxx = ut , 0 < x < L, t > 0, where u(x, t)
represents the temperature at time t and
position x along the rod with initial and
boundary conditions discussed.
 Method of separation of variables to
solve the heat equation is discussed.

Section 9.7 The Wave Equation:
Vibrations of an Elastic String
The classical wave equation that describes
the vibrations of an elastic string of length L
is a2uxx = utt, 0 < x < L, t > 0, where u(x, t)
represents the string’s displacement from
equilibrium at time t and position x along
the string.
 Initial and boundary conditions are given.
 Solved by method of separation of
variables.
 Defined modal shape and natural
frequency.

Section 9.8 Laplace’s Equation
9.8 Ctd. Dirichlet problem